Answer:
m = 4.7 μg
Explanation:
Given data:
density of acetone = 60.0 μg/L
Volume = 79.0 mL
Mass = ?
Solution:
Formula:
d = m/v
v = 79.0 mL × 1L /1000 mL
v = 0.079 L
Now we will put the values on formula:
d = m/v
60.0 μg/L = m/0.079 L
m = 60.0 μg/L × 0.079 L
m = 4.7 μg
So health risk limit for acetone = 4.7 μg
Answer:
0.15g
Explanation:
Given parameters:
Number of molecules of water = 1.2 x 10²¹ molecules
Unknown:
Mass of SnO₂ = ?
Solution:
To solve this problem, we have to work from the known to the unknown specie;
SnO₂ + 2H₂ → Sn + 2H₂O
Ensure that the equation given is balanced;
Now,
the known species is water;
6.02 x 10²³ molecules of water = 1 mole
1.2 x 10²¹ molecules of water =
= 0.2 x 10⁻²moles
Number of moles of water = 0.002moles
From the balanced chemical equation:
2 mole of water is produced from 1 mole of SnO₂
0.002 moles of water will be produced from
= 0.001moles
To find the mass;
Mass = number of moles x molar mass
Molar mass of SnO₂ = 118.7 + 2(16) = 150.7g/mol
Mass = 0.001 x 150.7 = 0.15g
Atoms are (stable) if it has (valence) electrons in its outer most energy level.
The amount of heat released by the sample has been 22.54 kJ. Thus, option C is correct.
The specific heat has been defined as the amount of heat required to raise the temperature of 1 gram of substance by 1 degree Celsius.
The specific heat has been expressed as:

<h3 /><h3>Computation for the heat absorbed</h3>
The iron and calorimeter are in side the closed system. Thus, the energy released by the sample, has been equivalent to the energy absorbed by the calorimeter.

The given mass of calorimeter has been, 
The specific heat of the calorimeter has been, 
The change in temperature of the calorimeter has been, 
Substituting the values for heat released:

The amount of heat released by the sample has been 22.54 kJ. Thus, option C is correct.
Learn more about specific heat, here:
brainly.com/question/2094845
Based on the given conditions, the gravitational force acted on the body is greater than the force upward by a quantity equal to 500 N. Therefore, the net force will cancel out and will remain an amount of 500N downward relative to the gravitational force exerted by the object.